Question

What is `DeltaH_f^@` for ammonia, and how is it formulated?

Answer 1

The `"standard enthalpy of formation"` of ammonia is `-46*kJ*mol^-1`.

Explanation:

What does this mean...? Well the given equation is associated with the RELEASE of a given quantity of heat to the surroundings........

`1/2N_2(g) + 3/2H_2(g) rarr NH_3(g) +46*kJ`

We write enthalpies (and other thermodynamic quantities) `"per mole of reaction"` as written. And thus when we report `DeltaH_"rxn"^@=DeltaH_f^@(NH_3)=-46*kJ*mol^-1`......The negative sign indicates that heat is RELEASED to the environment rather than absorbed. The enthalpy of the `3xxNH` formed is GREATER than the enthalpy of the `H-H` and `N-=N` bonds broken.

And so if there are `2*mol` ammonia formed, i.e. `34*g` rather than `17*g`, the enthalpy is DOUBLED......... Are you clear on what I am saying?